Linear Algebra: Basis, Dimension, and Subspace of 2x2 Matrices

Published on July 6, 2025

Linear-Algebra

This linear algebra problem involves determining the basis and dimension of a 2x2 matrix space and a subspace defined by linear equations. The solution involves finding the inverse of a matrix and verifying subspace properties.

Question

Answer

Here’s a step-by-step solution to the problem.

Part 1: Properties of

1.a) Determine the basis and dimension of .

The set

consists of all

matrices with real entries. A general matrix in

can be written as:

where

We can express this matrix as a linear combination of simpler matrices:

The set of matrices

forms a basis for

because:

Spanning Set: Any matrix in can be written as a linear combination of these four matrices.
Linear Independence: If , then , which implies .

Since there are 4 matrices in the basis, the dimension of

is 4.

Basis:

Dimension: 4

Part 2: Properties of the Subspace

The set

is defined as:

From the given conditions:

The variable

is not constrained by

. So

can be any real number.
Thus, a matrix in

can be written as:

2.a) Give an example of matrix in such that all entries in are non-zero.

For all entries to be non-zero, we need

, and

.
Since

and

, if

, then

and

will also be non-zero.
Let’s choose a value for

that makes the fractions easy to work with, for example, a multiple of 21 (LCM of 7 and 3). Let

.
Then:

Now, we need to choose

. Let

So, an example matrix

is:

All entries (21, -6, 1, -35) are non-zero.

2.b) Find the inverse of matrix .

For a

matrix

, its inverse is

, provided

.
For

Since

, the inverse exists.

We can simplify the fractions:

So,

2.c) Is a subspace of ? If yes, verify all 3 conditions of subspace are satisfied.

To be a subspace,

must satisfy three conditions:

Non-empty (contains the zero vector): The zero matrix must be in .
For , we have .
Check the conditions:
. (Satisfied)
. (Satisfied)
Since the zero matrix satisfies both conditions, is non-empty.
Closure under addition: If and , then .
Let and .
Since :

Since :

Consider .
Check the conditions for :
. (Satisfied)
. (Satisfied)
Thus, .
Closure under scalar multiplication: If and , then .
Let . So:

Consider .
Check the conditions for :
. (Satisfied)
. (Satisfied)
Thus, .